TITLE: How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$
QUESTION [5 upvotes]: How does $e^0 = 1$ if you define $e^x = \sum_{n = 0}^\infty x^n/n!$ since $e^0 = 0^0$, and we know the right hand side is undefined?

REPLY [5 votes]: In this situation, you should view that $0^0$ as
$$\lim_{x \to 0} x^0,$$
which is pretty easily seen to be $1$.
In situations of continuity, it is often better do define $0^0$ as $1$.  One reason I've heard for this is that continuity of $x^0$ is more useful than continuity of the related function $0^x$.